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spherical geometry (was Re: navigation riddle)

Regarding Joel R's comment:

For me, the impediment is partially that; but more so that I've
heard this one since I was a kid and was always told the answer was
the North Pole (combined probably with a northern hemisphere
prejudice) all of this combines in my head to provide a powerful
disinclination to search for other answers beyond the one given.

As an aside, both answers provide a nice example of how geometry on
S2 (surface of a 3dim sphere) is different from geometry on R2
(Euclidean plane).

Joel R

This discussion reminds me of a somewhat related problem. The
Northern Hemisphere solution can be considered as a closed path that
is, to a good approximation, a closed circular sector whose wedge
angle is 1 radian. The 2nd leg of the path is a circular arc of
approximately 1 radian in turning angle and the 1st and 3rd legs of
the path are two (geodesically straight) radii connecting the arc
ends to the center point of the arc.

Suppose we straightened out the 2nd leg of the path so *all three*
legs of the path are geodesically straight and the length of the 2nd
leg is the same as the length of the 1st and 3rd legs. The whole
closed path is now an equilateral triangle as inscribed onto the
spherical surface. The problem is to find a formula for the measure
of the interior angle of such an equilateral triangle as a function
of the length s of the sides of the triangle (conveniently in units
of the sphere's radius). A few hints are that 1) the value of the
formula must boil down to 60 deg in the limit of s becoming a
zeroth fraction of the sphere's radius, 2) the value of the formula
becomes 90 deg when s is 1/4 of the circumference of the sphere,
3) the maximum size triangle occurs for a great circle with 3
equally-spaced vertices (120 deg apart from each other) on it with
the interior angle at each vertex being 180 deg across the vertex and
each side having a length s of 1/3 of the sphere's circumference, and
4) the messy intermediate math eventually simplifies to a relatively
simplified formula in the general case.

For a lot of extra credit points you can also find the proper formula
for the *area* of this spherical equilateral triangle in terms of the
length s of the sides of the triangle (making sure that the formula
boils down to all the correct values for the variously known special

David Bowman